已知函数y=1 2sin(2x π 6) 5 4,-搜答案-智慧作业帮

分析:函数y=2sin(2x-π/4)的图象的对称轴的位置为取最值的地方,对称中心为函数值为0的地方.因为2x-π/4=kπ+π/2(k为整数)解得x=kπ/2+3π/8,所以函数y=2sin(2x-

y=2sin(2x+π/3)由函数可以看出：振幅：A=2周期：T=(2π)/2=π初相：φ=π/3(x=0时)

1A=2T=2π/2=π初相=π/322x+π/3=π/2+kπ所以对称轴为x=π/12+kπ/2-π/2+2kπ≦2x+π/3≦π/2+2kπ所以-5π/12+kπ≦x≦π/12+kπ所以递增区间为

振幅为2；周期为π；初相为π/3单增区间：kπ－5π／12≦x≦kπ＋π／12对称轴：x=﹙1／2﹚kπ+（1／12）π

（0,1）代入原式知sinφ=1/2因为|φ|

最大值 5 ,此时sin(2x-π/3)=1,2x-π/3=π/2+2Kπ,得X=5π/12+Kπ最小值 51,此时sin(2x-π/3)=-1,2x-π/3=-π/2+

解由x属于[π/12,π/3]即π/12≤x≤π/3即π/6≤2x≤2π/3即-π/6≤2x-π/3≤π/3即-1/2≤sin（2x-π/3）≤√3/2即-1≤2sin（2x-π/3）≤√3即-1≤y

由化简sinx+cosx前分别乘以根号2*sin45.根号2*cos45.,得解根号2sinxy=sinx的平方+根好2*sinx+2令t=sinx-1=

y=sin²x+sinx+cosx+2=(1-cos2x)/2+√2sin(x+л/4)+2=(1/2)*sin(2x+л/2)+√2*sin(x+л/4)+5/2；=(1/2)*sin(2

我列个去,就算我高中毕业到现在已经8年了,我也看的出来1楼的乱说的撒,值域明显是[-2,2]嘛

(1)x-π/12π/65π/122π/311π/122x+π/60π/2π3π/22πy=1/22sin(2x+π/6)01/20-1/20(2)由题意,A=1/2设最小正周期为T,则T/2=4π/

(-π/2,π/2)应小于等于半个周期,.-1≤ω≤1,又函数是减函数,sin(-ωπ/2)>sin(ωπ/2),sin(ωπ/2)

y=2cosxsin(x+π/3)-根号3*(sin^2)x+sinxcosx,后两项先提出一个sinx,然后括号内部分用叠加原理,得到y=2cosxsin(x+π/3)+2sinxcos(x+π/3

因为,-π/2

x∈[-2π/9,π/6]3x+π/3∈[-π/3,5π/6]sin(3x+π/3)∈[-√3/2,1]2sin(3x+π/3)∈[-√3,2]函数的最大值=2函数的最小值=-√3

原式=(1-cos2x)/2+(sin2x)/2+2=(sin2x-cos2x)/2+5/2=(sin（2x-45度）)*(根号2)/2+5/2所以是大于（根号2+5）/2,小于（5-根号2）/2

解1当2kπ-π/2≤2x+π/3≤2kπ+π/2,k属于Z时,y是增函数即2kπ-5π/6≤2x≤2kπ+π/6,k属于Z时,y是增函数即kπ-5π/12≤x≤kπ+π/12,k属于Z时,y是增函数

函数y=2sin(3x+π/6)当函数y取最大值时有3x+π/6=2kπ+π/2即x=2kπ/3+π/9,k∈Z所以x得集合为{x|x=2kπ/3+π/9,k∈Z}

1:y=2sin(x+π/6）-2cosx=2[sinxcosπ/6+cosxsinπ/6]-2cosx=√3sinx+cosx-2cosx=√3sinx-cosx=2sin(x-π/6）2:y=2c